3.147 \(\int \frac {x^2}{\sin ^{-1}(a+b x)^2} \, dx\)

Optimal. Leaf size=84 \[ -\frac {\left (4 a^2+1\right ) \text {Si}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {2 a \text {Ci}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}+\frac {3 \text {Si}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {x^2 \sqrt {1-(a+b x)^2}}{b \sin ^{-1}(a+b x)} \]

[Out]

-2*a*Ci(2*arcsin(b*x+a))/b^3-1/4*(4*a^2+1)*Si(arcsin(b*x+a))/b^3+3/4*Si(3*arcsin(b*x+a))/b^3-x^2*(1-(b*x+a)^2)
^(1/2)/b/arcsin(b*x+a)

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Rubi [A]  time = 0.23, antiderivative size = 161, normalized size of antiderivative = 1.92, number of steps used = 12, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {4805, 4745, 4621, 4723, 3299, 4631, 3302} \[ -\frac {a^2 \text {Si}\left (\sin ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {2 a \text {CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Si}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {3 \text {Si}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {2 a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSin[a + b*x]^2,x]

[Out]

-((a^2*Sqrt[1 - (a + b*x)^2])/(b^3*ArcSin[a + b*x])) + (2*a*(a + b*x)*Sqrt[1 - (a + b*x)^2])/(b^3*ArcSin[a + b
*x]) - ((a + b*x)^2*Sqrt[1 - (a + b*x)^2])/(b^3*ArcSin[a + b*x]) - (2*a*CosIntegral[2*ArcSin[a + b*x]])/b^3 -
SinIntegral[ArcSin[a + b*x]]/(4*b^3) - (a^2*SinIntegral[ArcSin[a + b*x]])/b^3 + (3*SinIntegral[3*ArcSin[a + b*
x]])/(4*b^3)

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 4621

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^(n + 1))
/(b*c*(n + 1)), x] + Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSin[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Free
Q[{a, b, c}, x] && LtQ[n, -1]

Rule 4631

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcSin
[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 4745

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d + e
*x)^m*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {x^2}{\sin ^{-1}(a+b x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2}{b^2 \sin ^{-1}(x)^2}-\frac {2 a x}{b^2 \sin ^{-1}(x)^2}+\frac {x^2}{b^2 \sin ^{-1}(x)^2}\right ) \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {x}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}+\frac {\operatorname {Subst}\left (\int \left (-\frac {\sin (x)}{4 x}+\frac {3 \sin (3 x)}{4 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \sin ^{-1}(x)} \, dx,x,a+b x\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {2 a \text {Ci}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}-\frac {\operatorname {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\sin (3 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a^2 \operatorname {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {2 a \text {Ci}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Si}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a^2 \text {Si}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac {3 \text {Si}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 86, normalized size = 1.02 \[ -\frac {\frac {4 b^2 x^2 \sqrt {-a^2-2 a b x-b^2 x^2+1}}{\sin ^{-1}(a+b x)}+\left (4 a^2+1\right ) \text {Si}\left (\sin ^{-1}(a+b x)\right )+8 a \text {Ci}\left (2 \sin ^{-1}(a+b x)\right )-3 \text {Si}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSin[a + b*x]^2,x]

[Out]

-1/4*((4*b^2*x^2*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2])/ArcSin[a + b*x] + 8*a*CosIntegral[2*ArcSin[a + b*x]] + (1
+ 4*a^2)*SinIntegral[ArcSin[a + b*x]] - 3*SinIntegral[3*ArcSin[a + b*x]])/b^3

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{\arcsin \left (b x + a\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arcsin(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x^2/arcsin(b*x + a)^2, x)

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giac [B]  time = 0.39, size = 169, normalized size = 2.01 \[ -\frac {a^{2} \operatorname {Si}\left (\arcsin \left (b x + a\right )\right )}{b^{3}} - \frac {2 \, a \operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{3}} + \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a}{b^{3} \arcsin \left (b x + a\right )} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a^{2}}{b^{3} \arcsin \left (b x + a\right )} + \frac {3 \, \operatorname {Si}\left (3 \, \arcsin \left (b x + a\right )\right )}{4 \, b^{3}} - \frac {\operatorname {Si}\left (\arcsin \left (b x + a\right )\right )}{4 \, b^{3}} + \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{b^{3} \arcsin \left (b x + a\right )} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{b^{3} \arcsin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arcsin(b*x+a)^2,x, algorithm="giac")

[Out]

-a^2*sin_integral(arcsin(b*x + a))/b^3 - 2*a*cos_integral(2*arcsin(b*x + a))/b^3 + 2*sqrt(-(b*x + a)^2 + 1)*(b
*x + a)*a/(b^3*arcsin(b*x + a)) - sqrt(-(b*x + a)^2 + 1)*a^2/(b^3*arcsin(b*x + a)) + 3/4*sin_integral(3*arcsin
(b*x + a))/b^3 - 1/4*sin_integral(arcsin(b*x + a))/b^3 + (-(b*x + a)^2 + 1)^(3/2)/(b^3*arcsin(b*x + a)) - sqrt
(-(b*x + a)^2 + 1)/(b^3*arcsin(b*x + a))

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maple [A]  time = 0.12, size = 149, normalized size = 1.77 \[ \frac {-\frac {a \left (2 \Ci \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-\sin \left (2 \arcsin \left (b x +a \right )\right )\right )}{\arcsin \left (b x +a \right )}-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{4 \arcsin \left (b x +a \right )}-\frac {\Si \left (\arcsin \left (b x +a \right )\right )}{4}+\frac {\cos \left (3 \arcsin \left (b x +a \right )\right )}{4 \arcsin \left (b x +a \right )}+\frac {3 \Si \left (3 \arcsin \left (b x +a \right )\right )}{4}-\frac {a^{2} \left (\Si \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{\arcsin \left (b x +a \right )}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arcsin(b*x+a)^2,x)

[Out]

1/b^3*(-a*(2*Ci(2*arcsin(b*x+a))*arcsin(b*x+a)-sin(2*arcsin(b*x+a)))/arcsin(b*x+a)-1/4/arcsin(b*x+a)*(1-(b*x+a
)^2)^(1/2)-1/4*Si(arcsin(b*x+a))+1/4/arcsin(b*x+a)*cos(3*arcsin(b*x+a))+3/4*Si(3*arcsin(b*x+a))-a^2*(Si(arcsin
(b*x+a))*arcsin(b*x+a)+(1-(b*x+a)^2)^(1/2))/arcsin(b*x+a))

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arcsin(b*x+a)^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{{\mathrm {asin}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/asin(a + b*x)^2,x)

[Out]

int(x^2/asin(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\operatorname {asin}^{2}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/asin(b*x+a)**2,x)

[Out]

Integral(x**2/asin(a + b*x)**2, x)

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